but i cannot understand how the radius does not depend on the length of the cord or the length of the top, horizontal bar??

The radius does depend on those quantities, but their value is encoded in the value of [itex]\alpha[/itex]. If [itex]\omega[/itex] is fixed and we increase the length of the cord then [itex]\alpha[/itex] increases (and if we decrease the length of the cord it decreases). Similarly, for a fixed [itex]\omega[/itex], an increase in the length of the horizontal bar will increase [itex]\alpha[/itex]. The physics of what is happening (spinning a mass on a string) means that the value of [itex]\alpha[/itex] has a dependency on these quantities.

If you wanted, you could make this dependency explicit. From the geometry of the problem you can show that,

where [itex]a[/itex] is the length of the horizontal bar and [itex]b[/itex] is the length of the cord. If you substitute this into your answer then you can get an expression for [itex]R[/itex] that depends explicitly on [itex]a[/itex] and [itex]b[/itex] instead of [itex]\alpha[/itex]. However rearranging the answer involves solving a quartic equation for [itex]R[/itex] which is not easy.

As an interesting aside, note that understanding the physics doesn't give us any 'information for free'. If we are given [itex]\alpha[/itex] then to find [itex]R[/itex] geometrically we require two more pieces of information: the length of the horizontal bar and the length of the string. To find [itex]R[/itex] from [itex]\alpha[/itex] using instead the physics of the situation, we still require two more pieces of information (the speed of rotation [itex]\omega[/itex] and the downward acceleration the ball would experience if removed from the string, [itex]g[/itex]).